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Theorem sbc3or 41840
Description: sbcor 3756 with a 3-disjuncts. This proof is sbc3orgVD 42159 automatically translated and minimized. (Contributed by Alan Sare, 31-Dec-2011.) (Revised by NM, 24-Aug-2018.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sbc3or ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3or
StepHypRef Expression
1 sbcor 3756 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ ([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒))
2 df-3or 1090 . . . . 5 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
32bicomi 227 . . . 4 (((𝜑𝜓) ∨ 𝜒) ↔ (𝜑𝜓𝜒))
43sbcbii 3764 . . 3 ([𝐴 / 𝑥]((𝜑𝜓) ∨ 𝜒) ↔ [𝐴 / 𝑥](𝜑𝜓𝜒))
5 sbcor 3756 . . . 4 ([𝐴 / 𝑥](𝜑𝜓) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓))
65orbi1i 914 . . 3 (([𝐴 / 𝑥](𝜑𝜓) ∨ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
71, 4, 63bitr3i 304 . 2 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
8 df-3or 1090 . 2 (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓) ∨ [𝐴 / 𝑥]𝜒))
97, 8bitr4i 281 1 ([𝐴 / 𝑥](𝜑𝜓𝜒) ↔ ([𝐴 / 𝑥]𝜑[𝐴 / 𝑥]𝜓[𝐴 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wo 847  w3o 1088  [wsbc 3703
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-8 2113  ax-9 2121  ax-10 2142  ax-12 2176  ax-ext 2709
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3or 1090  df-tru 1546  df-ex 1788  df-nf 1792  df-sb 2072  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3417  df-sbc 3704
This theorem is referenced by:  sbcoreleleq  41843
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